如何从数据库中获取项目的标题并将其发送到CodeIgniter中的标题模板

Question

我正在CodeIgniter中编写一个应用程序,我<title>在每个控制器的每个页面上指定元标记,我已设法将其发送到我的标题模板.但是,现在我已经创建了一个应用程序,通过CodeIgniter模型从数据库中获取信用卡及其标题.我想自动获取并使用信用卡的名称,<title>这样我就不需要手动更改它了,但我有点不知道如何继续.

这是我现在的代码:

调节器

public function show($card = NULL)
{

    $data['query'] = $this->Listing_model->get_card($card);

    $header["page_title"] = from the model

    $this->load->view('includes/header',$header);
    $this->load->view('listings/listing_card',$data);
    $this->load->view('includes/footer');
}

模型

function get_card($card = FALSE)
{
    $query = $this->db->get_where('creditcards', array('slug' => $card), 0,1);
    return $query->result();
}

我在创建这个应用程序时一直遵循官方的CodeIgniter文档,但到目前为止还没有运气.有解决方案吗

Answer 1

试试这个

1. 模型改变了
2. 控制器已更改.

在模型中

function get_card($card)
{
    $query = $this->db->query("SELECT * FROM table_name WHERE creditcards = '$card' ");
    $result = $query->result_array();
    $count = count($result); # New

    if(empty($count)){ # New
        return FALSE;
    }
    elseif($count > 1){ # New
        return 0;
    }
    else{
        return $result;
    }
}

在控制器中

public function show($card)
{
    $result = $this->Listing_model->get_card($card); # Changed

    if($result == FALSE){ # New
        echo "No Data Found";
    }
    elseif($result == 0){ # New
        echo "Multiple Data Found";
    }
    else{
        $data["page_title"] = $result[0]['field_name']; # Changed

        $this->load->view('includes/header',$data); # Changed
        $this->load->view('listings/listing_card',$data);
        $this->load->view('includes/footer');
    }

}

在视图中

<?php echo (!empty($page_title)) ? $page_title : ''; ?> # Changed 

Answer 2

创建一个基本控制器.默认位置是application/core/MY_Controller.php =>可以通过配置更改.通过使用,$this->site_data您可以在基类中添加变量,并在每个控件/视图中使用它们

class MY_Controller extends CI_Controller
{
    function __construct()
    {
        parent::__construct();


        $this->load->database();

        $this->load->model('your model');

        $result = $this->Listing_model->get_card($card);
        $this->site_data['query']=$result;

       $this->site_data_header["page_title"] = $result['get the property you want'];//this is possible, as get_card returns 1 result 


    }
}


class YourClass extends MY_Controller
{
    function __construct()
    {
        parent::__construct();
    }

    public function show($card = NULL)
    {
        //you don't need to split the variables 
        //for header and the rest 

        $this->load->view('includes/header',$this->site_data_header);
        $this->load->view('listings/listing_card',$this->site_data);
        $this->load->view('includes/footer');
    }
}

我认为你的get_where错了:

$query = $this->db->get_where('mytable', array('id' => $id), $limit, $offset);

你的限制是0

function get_card($card = FALSE)
{
    $query = $this->db->get_where('creditcards', array('slug' => $card), 1,0);//limit 1 offset 0
    return $query->result();
}

访问视图中的数据

<title><?php echo (!isset($page_title) ? '' : $page_title) ?></title>